3.1 \(\int \frac {\sec ^{-1}(a x^5)}{x} \, dx\)

Optimal. Leaf size=62 \[ \frac {1}{10} i \text {Li}_2\left (-e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right ) \]

[Out]

1/10*I*arcsec(a*x^5)^2-1/5*arcsec(a*x^5)*ln(1+(1/a/x^5+I*(1-1/a^2/x^10)^(1/2))^2)+1/10*I*polylog(2,-(1/a/x^5+I
*(1-1/a^2/x^10)^(1/2))^2)

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Rubi [A]  time = 0.09, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5218, 4626, 3719, 2190, 2279, 2391} \[ \frac {1}{10} i \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a*x^5]/x,x]

[Out]

(I/10)*ArcSec[a*x^5]^2 - (ArcSec[a*x^5]*Log[1 + E^((2*I)*ArcSec[a*x^5])])/5 + (I/10)*PolyLog[2, -E^((2*I)*ArcS
ec[a*x^5])]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5218

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCos[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^{-1}\left (a x^5\right )}{x} \, dx &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {\sec ^{-1}(a x)}{x} \, dx,x,x^5\right )\\ &=-\left (\frac {1}{5} \operatorname {Subst}\left (\int \frac {\cos ^{-1}\left (\frac {x}{a}\right )}{x} \, dx,x,\frac {1}{x^5}\right )\right )\\ &=\frac {1}{5} \operatorname {Subst}\left (\int x \tan (x) \, dx,x,\sec ^{-1}\left (a x^5\right )\right )\\ &=\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {2}{5} i \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}\left (a x^5\right )\right )\\ &=\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{5} \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}\left (a x^5\right )\right )\\ &=\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )-\frac {1}{10} i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}\left (a x^5\right )}\right )\\ &=\frac {1}{10} i \sec ^{-1}\left (a x^5\right )^2-\frac {1}{5} \sec ^{-1}\left (a x^5\right ) \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\frac {1}{10} i \text {Li}_2\left (-e^{2 i \sec ^{-1}\left (a x^5\right )}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.90 \[ \frac {1}{10} i \left (\text {Li}_2\left (-e^{2 i \sec ^{-1}\left (a x^5\right )}\right )+\sec ^{-1}\left (a x^5\right ) \left (\sec ^{-1}\left (a x^5\right )+2 i \log \left (1+e^{2 i \sec ^{-1}\left (a x^5\right )}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a*x^5]/x,x]

[Out]

(I/10)*(ArcSec[a*x^5]*(ArcSec[a*x^5] + (2*I)*Log[1 + E^((2*I)*ArcSec[a*x^5])]) + PolyLog[2, -E^((2*I)*ArcSec[a
*x^5])])

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fricas [F]  time = 2.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcsec}\left (a x^{5}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a*x^5)/x,x, algorithm="fricas")

[Out]

integral(arcsec(a*x^5)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsec}\left (a x^{5}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a*x^5)/x,x, algorithm="giac")

[Out]

integrate(arcsec(a*x^5)/x, x)

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arcsec}\left (a \,x^{5}\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(a*x^5)/x,x)

[Out]

int(arcsec(a*x^5)/x,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -5 \, a^{2} \int \frac {\sqrt {a x^{5} + 1} \sqrt {a x^{5} - 1} \log \relax (x)}{a^{4} x^{11} - a^{2} x}\,{d x} - 5 i \, a^{2} \int \frac {\log \relax (x)}{a^{4} x^{11} - a^{2} x}\,{d x} + \arctan \left (\sqrt {a x^{5} + 1} \sqrt {a x^{5} - 1}\right ) \log \relax (x) - \frac {1}{2} i \, \log \left (a^{2} x^{10}\right ) \log \relax (x) + \frac {1}{2} i \, \log \left (a x^{5} + 1\right ) \log \relax (x) + \frac {1}{2} i \, \log \left (-a x^{5} + 1\right ) \log \relax (x) + i \, \log \relax (a) \log \relax (x) + \frac {5}{2} i \, \log \relax (x)^{2} + \frac {1}{10} i \, {\rm Li}_2\left (a x^{5}\right ) + \frac {1}{10} i \, {\rm Li}_2\left (-a x^{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(a*x^5)/x,x, algorithm="maxima")

[Out]

-5*a^2*integrate(sqrt(a*x^5 + 1)*sqrt(a*x^5 - 1)*log(x)/(a^4*x^11 - a^2*x), x) - 5*I*a^2*integrate(log(x)/(a^4
*x^11 - a^2*x), x) + arctan(sqrt(a*x^5 + 1)*sqrt(a*x^5 - 1))*log(x) - 1/2*I*log(a^2*x^10)*log(x) + 1/2*I*log(a
*x^5 + 1)*log(x) + 1/2*I*log(-a*x^5 + 1)*log(x) + I*log(a)*log(x) + 5/2*I*log(x)^2 + 1/10*I*dilog(a*x^5) + 1/1
0*I*dilog(-a*x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (\frac {1}{a\,x^5}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/(a*x^5))/x,x)

[Out]

int(acos(1/(a*x^5))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asec}{\left (a x^{5} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(a*x**5)/x,x)

[Out]

Integral(asec(a*x**5)/x, x)

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